3.II.15H

Methods | Part IB, 2005

Obtain the power series solution about t=0t=0 of

(1t2)d2 dt2y2tddty+λy=0\left(1-t^{2}\right) \frac{\mathrm{d}^{2}}{\mathrm{~d} t^{2}} y-2 t \frac{\mathrm{d}}{\mathrm{d} t} y+\lambda y=0

and show that regular solutions y(t)=Pn(t)y(t)=P_{n}(t), which are polynomials of degree nn, are obtained only if λ=n(n+1),n=0,1,2,\lambda=n(n+1), n=0,1,2, \ldots Show that the polynomial must be even or odd according to the value of nn.

Show that

11Pn(t)Pm(t)dt=knδnm\int_{-1}^{1} P_{n}(t) P_{m}(t) \mathrm{d} t=k_{n} \delta_{n m}

for some kn>0k_{n}>0.

Using the identity

(x2x2x+t(1t2)t)1(12xt+x2)12=0,\left(x \frac{\partial^{2}}{\partial x^{2}} x+\frac{\partial}{\partial t}\left(1-t^{2}\right) \frac{\partial}{\partial t}\right) \frac{1}{\left(1-2 x t+x^{2}\right)^{\frac{1}{2}}}=0,

and considering an expansion nan(x)Pn(t)\sum_{n} a_{n}(x) P_{n}(t) show that

1(12xt+x2)12=n=0xnPn(t),0<x<1\frac{1}{\left(1-2 x t+x^{2}\right)^{\frac{1}{2}}}=\sum_{n=0}^{\infty} x^{n} P_{n}(t), \quad 0<x<1

if we assume Pn(1)=1P_{n}(1)=1.

By considering

11112xt+x2dt\int_{-1}^{1} \frac{1}{1-2 x t+x^{2}} d t

determine the coefficient knk_{n}.

Typos? Please submit corrections to this page on GitHub.